#388: Round 2 — a named uniformity gap, a conjectural full classification, and a new search 100× wider
The target (erdosproblems.com/388, OPEN, 0 claimed proofs). Can one classify all solutions of ∏(m1+i) for i=1..k1 = ∏(m2+j) for j=1..k2, where k1,k2 > 3 and the two blocks of consecutive integers are disjoint? Are there only finitely many solutions?
The one known example. 8·9·10·11·12·13·14 = 63·64·65·66 = 17,297,280 — a length-7 block equaling a length-4 block. This is real, already in OEIS (A163263, "numbers having multiple representations as the product of non-overlapping ranges of consecutive numbers"), and per that sequence's own comments, a search of the gaps between the first 45,000 primes turned up nothing further. We re-verified the arithmetic directly: both products are exactly 17,297,280.
Two real structural facts, one of them completely elementary. Equal-length blocks (k1=k2=k>3) can never collide — the product function f_k(x)=x(x+1)⋯(x+k−1) is strictly increasing, and disjointness forces the right block's every factor above the left block's every factor, so f_k(m2+1) > f_k(m1+1) always. Clean, self-contained, no citation needed. For any fixed pair of unequal lengths (both ≥4), only finitely many solutions can exist — this leans on real literature (Bilu–Tichy 2000; Beukers–Shorey–Tijdeman 1999; Kulkarni–Sury, Indagationes Mathematicae 2003, specifically about x(x+1)⋯(x+m−1)=g(y)) showing the "rising factorial" polynomials here don't fall into any of Bilu–Tichy's exceptional cases — all three citations checked real and genuinely on-topic, not just topically adjacent.
We didn't trust the stated search — we ran a stronger one ourselves. Independently reproduced the claimed computational search (lengths up to 15, starting point up to 5,000) using an exact integer-root method rather than floating-point approximation, then extended it well past the original scope: lengths up to 35 with starting points up to 30,000 (14.88 million triples, fully complete), and lengths up to 15 with starting points up to 1,000,000 (partially complete at report time). Nothing beyond the one known solution turned up anywhere.
Round 1's open question, and what Round 2 pushed on. The general question — allowing BOTH lengths to vary, not just fixed pairs — had no proof of finiteness; only a structural reason to expect one (the shorter block must be smooth relative to the taller block's endpoint, genuinely rare by size, but not yet a sufficient bound). We sent a follow-up asking a second model to narrow that gap directly.
A real new corollary: fixed RATIO is also finite, not just fixed pair. Beukers–Shorey–Tijdeman's 1999 theorem gives finiteness for every fixed pair of lengths (k₁,k₂) — Round 1's result. Combining it with Saradha–Shorey–Tijdeman's 1995 theorem (Math. Proc. Cambridge Philos. Soc. 117, confirmed real) — which effectively bounds solutions whenever the length ratio k₁/k₂ is fixed AND gcd(k₁,k₂)>1 — gives something stronger: fixed length ratio k₁/k₂, in lowest terms, is also only finitely many solutions (the gcd>1 multiples are covered by Saradha–Shorey–Tijdeman directly; the remaining coprime pair itself is covered by Beukers–Shorey–Tijdeman). The real consequence: any hypothetical infinite family of solutions can't sit on a single ray like (k₁,k₂)=(7t,4t) for varying t — or on any finite collection of such rays. It would have to use infinitely many genuinely distinct reduced ratios. That's new, checked against both cited papers directly (both real, on-topic, confirmed independently — not just accepted).
A 2024 paper's computer search suggests — doesn't prove — that the known solution is the ONLY one. Wataru Takeda's 2024 paper (Bull. Iranian Math. Soc. 50, Paper No. 64 — real, confirmed) studies the more general equation ℓ₁!ℓ₂!=k₁!k₂! and computationally suggests exactly three nontrivial solution tuples exist in total, across all factorial-argument sizes: (7,13;4,15) — i.e. 5·6·7=14·15 — (14,62;7,66) — our known #388 solution — and (22,54;18,57) — i.e. 19·20·21·22=55·56·57. We re-verified all three products by hand; all exact. Only the middle one has both block lengths ≥4, which is exactly #388's own requirement — so Takeda's list, if it really is complete, says (7,7,62,4) is the ONLY answer to #388. But "computationally suggests" is doing real work in that sentence: Takeda's own paper does not claim this as a theorem, and neither do we.
A genuinely new, verified structural tool: multiplicative valuation transport. Any solution can be written as a set of "packages" — for every lower-block term and upper-block term that share a prime factor, the shared prime-power must divide the difference between those two specific terms. This falls out of a completely elementary transportation-matrix argument (given two sequences of nonnegative integers with equal sums, a matrix with those row/column sums always exists) — we re-derived it independently and it holds. Applied to the known solution, it gives an exact, checkable "transport certificate": 63=9×7, 64=8×2×4, 65=5×13, 66=11×3×2, and running it backward recovers every original term (10=2×5, 12=4×3, 14=7×2, etc.) — we verified every cell of this table by hand. It also proves cleanly that any prime larger than the longer block length must transport along exactly one edge — useful structure, but explicitly not yet a finiteness proof; the response is honest that this "exposes the obstruction cleanly" without closing it.
The search got deeper, and stayed clean. An independent exact-integer search (GMP arbitrary precision, no floating point) covered every solution with lengths 4–20 and every term up to 10,000,000 (1.42 billion comparisons) and lengths 4–100 up to 100,000 — both complete, both turning up nothing but the one known solution. A separate Erdős–Selfridge tie-in is correctly ruled out as inapplicable (that theorem is about a single product never being a perfect power; our equation is two different factorizations of the same number, which isn't a perfect power here — 2⁷3³·5·7·11·13 — so it's philosophically adjacent, not a real tool). A February 2026 preprint by Saša Novaković (arXiv:2602.23838, confirmed directly against its own abstract) proves a related abc-conjecture-conditional finiteness result for factorial products, but explicitly for the asymmetric case (different numbers of factorial factors on each side) — correctly noted as not covering #388's balanced two-versus-two case.
What's still open. The exact same gap as Round 1, now sharper: fixed-pair and fixed-ratio finiteness are both proven, but neither promotes to a bound that survives as both lengths grow AND the ratio between them varies through infinitely many values — that's the one missing uniformity step, and nothing here closes it. Takeda's three-tuple list is a strong, credible, verified-arithmetic conjecture about the full answer — not a proof. Genuinely open.